# Physical Limit to Building Size [duplicate]

This question already has an answer here:

Is there a limit as to how tall a building can theoretically* be? Why/Why not?

*By theoretically, I mean ignoring factors involving human discomfort (transportation, reduced air density affecting respiration, coldth etc), cost, material issues (Availability of substances etc).

I'm looking for purely physical answers with proper explanations.

## marked as duplicate by John Rennie, Jon Custer, innisfree, user36790, Qmechanic♦Oct 19 '16 at 9:24

• Well, if you are going to ignore materials issues, than I guess we can ignore many physical limits. So, And what is with all the bold words? What is a 'proper explanation' to you? – Jon Custer Oct 18 '16 at 16:26
• Does the building have to be on earth? – Floris Oct 18 '16 at 16:57
• @Floris I think I have just wrote the answer in your mind :) – Shing Oct 18 '16 at 17:35
• It largely depends on what you mean by "building". Is something similar to the Great Pyramid of Giza considered a "building" (basically, a huge pile of blocks of some material)? – valerio Oct 18 '16 at 18:17
• @Jon Before posting this question, I did a fair bit of research and glanced through multiple articles about why it's impossible, however, all of the factors listed boil down to cost and/or human discomfort. By a proper explanation, I mean one that list facts and doesn't theorize about how hard it would be construct a building. – Rippr Oct 19 '16 at 10:31

Yes. All materials will break under sufficient tension. And a sufficiently large building will experience a sufficiently large centrifugal force to rip it apart.

You may find the Wiki article on space elevators informative.

Yes. There is a limit. $\dagger$

Suppose the building is a cube with length $R$, then its weight $W$ is proportional to:

$W\propto R^3$

However the force $F$ provided by the crossection area to support its weight is:

$F\propto Area \propto R^2$

Eventually, there exist a $R$ such that

$R^3\gg R^2$

implies

$W\gg F$

Then the building will just collapse. Since near the bottom of the building will no loner in equilibrium at the direction of gravity, its own weight will make the bottom of the building break.

$\dagger$: Assuming you are talking about building on earth, also with a reasonable size related to Earth (I don't think something as huge as a Sun "on" earth should be considered as "building"), such that

1.) it is under a uniform gravity field;

2.) the rotational motion of earth can be ignored. $\dagger\dagger$

$\dagger\dagger$: However, even if these assumptions are not met, given a simple fact: stars all have a limited size. Our approximation does not seem too bad for $R >> R_{earth}$. (however, the physics reason is different for $R >> R_{earth}$)

• that is still a strength-of-material issue. what if the building was sculpted outa solid diamond? – robert bristow-johnson Oct 18 '16 at 18:15
• @robertbristow-johnson diamond has a finite breaking point even if it's a very high one, so I don't see how that would effect the above argument. – Virgo Oct 18 '16 at 18:24
• -1 Sufficiently large buildings experience enough centrifugal force (especially if they have a counterweight, see space elevators) to overcome this problem. So this does not pose a theoretical limit. – lemon Oct 18 '16 at 18:34
• @robertbristow-johnson It is not only a material issue. $W$ and $F$ depends on $r^2$ and $r^3$ respectively. Even if material matters (usually true), $r^3$ always wins out $r^2$. When it stands (not collapse yet), the material aspects win. but there always exists a $r$ such that the $r$ wins out the material – Shing Oct 18 '16 at 18:34
• @lemon rebuke accepted; and assumptions as well as explanations are added to the answer. If I have enough time, I might make a better argument. – Shing Oct 19 '16 at 16:35

Yes, there is something called square-cube law. You can read it on wikipedia

https://en.wikipedia.org/wiki/Square-cube_law

but basically, it explains why building taller and taller skyscrapers is increasingly difficult, and at some point it becomes impossible.

The basic idea is that consider you want to scale yourself to 10 of your size. Assuming that your density does not change, your mass, which is proportional to the cube of your size

$M \propto l^3$

your mass will be 1000 times larger. Now, your weight puts pressure on the cross section your femur bone (A). This pressure is given by

$P = M/A$

Thus, when your mass increases 1000 times, the cross section of your femur bone also has to go to 1000 times of itself, otherwise your bones will break. As the are is proportional to the square of the size of femur bone, your femur bone should scale 100 times of itself while your size only grows 10 times of itself. This means that as you grow in size, your bones get so thick that soon you will be noting but only a femur bone.

Similarly, if you want to scale a building 1000 times of itself, you have to make the building columns so thick that at some point, the building will be nothing but just a column.

There is a theory for glaciers. The maximum height is approximately $$h = \sqrt{\lambda L}$$ where $\lambda$ is proportional to the critical shear stress ($\lambda$ is about 10 metres for ice) and $L$ is half the width. It should apply to other materials, with a larger value of $\lambda$ for materials stronger than ice.